SOLUTION Water accelerated by a nozzle strikes the back surface of a cart moving horizontally at a constant velocity. The braking force, the power wasted by the brakes, and the acceleration of the cart if the brakes fail are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in all directions in the plane of the back surface. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 Fiction during motion is negligible. 5 The motions of the water jet and the cart are horizontal. 6 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, 𝛽 ≅ 1.
Analysis We take the cart as the control volume, and the direction of flow as the positive direction of x-axis. The relative velocity between the cart and the jet is
V_{r}=V_{\text {jet }}-V_{\text {cart }}=35-10=25 m / s
Therefore, we can view the cart as being stationary and the jet moving with a velocity of 25 m/s. Noting that water leaves the nozzle at 20 m/s and the corresponding mass flow rate relative to nozzle exit is 30 kg/s, the mass flow rate of water striking the cart corresponding to a water jet velocity of 25 m/s relative to the cart is
\dot{m}_{r}=\frac{V_{r}}{V_{\text {jet }}} \dot{m}_{\text {jet }}=\frac{25 m / s }{35 m / s }(30 kg / s )=21.43 kg / s
The momentum equation for steady flow in the x (flow)-direction reduces in this case to
\sum \vec{F}=\sum_{\text {out }} \beta \dot{m} \vec{V}-\sum_{\text {in }} \beta \dot{m} \vec{V} \rightarrow F_{R x}=-\dot{m}_{i} V_{i} \rightarrow F_{\text {brake }}=-\dot{m}_{r} V_{r}
We note that the brake force acts in the opposite direction to flow, and we should not forget the negative sign for forces and velocities in the negative x-direction. Substituting the given values,
F_{\text {brake }}=-\dot{m}_{r} V_{r}=-(21.43 kg / s )(+25 m / s )\left(\frac{1 N }{1 kg \cdot m / s ^{2}}\right)=-535.8 N \cong-536 N
The negative sign indicates that the braking force acts in the opposite direction to motion, as expected. Just as the water jet here imparts a force to the cart, the air jet from a helicopter (downwash) imparts a force on the surface of the water (Fig. 6–23). Noting that work is force times distance and the distance traveled by the cart per unit time is the cart velocity, the power wasted by the brakes is
\dot{W}=F_{\text {brake }} V_{\text {cart }}=(535.8 N )(10 m / s )\left(\frac{1 W }{1 N \cdot m / s }\right)=5358 W \cong 5.36 kW
Note that the power wasted is equivalent to the maximum power that can be generated as the cart velocity is maintained constant.
(c) When the brakes fail, the braking force will propel the cart forward, and the acceleration will be
a=\frac{F}{m_{\text {cart }}}=\frac{535.8 N }{400 kg }\left(\frac{1 kg \cdot m / s ^{2}}{1 N }\right)=1.34 m / s ^{2}
Discussion This is the acceleration at the moment the brakes fail. The acceleration will decrease as the relative velocity between the water jet and the cart (and thus the force) decreases.
